Question

A man of mass m is standing on one end of a plank of mass 2m floating on a river. The other end
just touches the bank of the river as shown in figure. With what minimum speed w.r.t. the plank
should the man jump to get out of the river. All surfaces are smooth and plank is always in level
with the bank. Given that the length of the plank is 15m.

(A) 7 m/s
(B) 7.5 m/s
(C) 9 m/s
(D) 5 m/s​

Answer 1

$\displaystyle\large\boxed {\sf {(C)\ 9\ m\ s^{-1}}}$

After the man reaches the other end of the plank, let the displacement of the man be $\displaystyle\sf {x_m}$ towards right and that of the plank be $\displaystyle\sf {x_p}$ towards left.

Then we have,

$\displaystyle\longrightarrow\sf {x_m+x_p=15\ m\quad\quad\dots (1)}$

The center of mass of the system,

$\displaystyle\longrightarrow\sf {\dfrac {m(x_m)+2m(-x_p)}{m+2m}=0}$

$\displaystyle\longrightarrow\sf {\dfrac {x_m-2x_p}{3}=0}$

$\displaystyle\longrightarrow\sf {x_m=2x_p}$

Then from (1),

$\displaystyle\longrightarrow\sf {x_p=5\ m}$

This means the river bank will be at a distance $\displaystyle\sf {5\ m}$ from the end of the plank.

For minimum speed of the man, the displacement of the plank is considered as the maximum horizontal range of the man, i.e.,

$\displaystyle\longrightarrow\sf {R_{max}=x_p}$

$\displaystyle\longrightarrow\sf {\dfrac {(v_m)^2}{g}=5\ m}$

$\displaystyle\longrightarrow\sf {v_m=7\ m\ s^{-1}}$

Well, the maximum horizontal range is considered for $\displaystyle\sf {\theta=45^{\circ}}$ where $\displaystyle\sf {\theta}$ is the angle made by his jumping with the horizontal.

Then, the velocity of the man with which the man jumps wrt ground,

$\displaystyle\longrightarrow\sf {\vec {v_m}=7\cos 45^{\circ}\hat i+7\sin 45^{\circ}\hat j}$

$\displaystyle\longrightarrow\sf {\vec{v_m}=\dfrac {7}{\sqrt2}\hat i+\dfrac {7}{\sqrt2}\hat j}$

Since the plank has no considerable vertical displacement nor vertical velocity, by law of conservation of linear momentum,

$\displaystyle\longrightarrow\sf {m(\vec {v_{m_x}})+2m(\vec {v_p})=0}$

$\displaystyle\longrightarrow\sf {\dfrac {7}{\sqrt2}\hat i+2(\vec {v_p})=0}$

$\displaystyle\longrightarrow\sf {\vec {v_p}=-\dfrac {7}{2\sqrt2}\hat i}$

where $\displaystyle\sf {\vec {v_p}}$ is the velocity of the plank wrt ground, acquired after the man jumps.

Then the magnitude of the relative velocity of the man wrt the plank is,

$\displaystyle\longrightarrow\sf {\left|\vec {v_{mp}}\right|=\left|\vec {v_m}-\vec {v_p}\right|}$

$\displaystyle\longrightarrow\sf {\left|\vec {v_{mp}}\right|=\left|\dfrac {7}{\sqrt2}\hat i+\dfrac {7}{\sqrt2}\hat j+\dfrac {7}{2\sqrt2}\hat i\right|}$

$\displaystyle\longrightarrow\sf {\left|\vec {v_{mp}}\right|=\left|\dfrac {21}{2\sqrt2}\hat i+\dfrac {7}{\sqrt2}\hat j\right|}$

$\displaystyle\longrightarrow\sf {\left|\vec {v_{mp}}\right|=\sqrt{\left(\dfrac {21}{2\sqrt2}\right)^2+\left (\dfrac {7}{\sqrt2}\right)^2}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {\left|\vec {v_{mp}}\right|=9\ m\ s^{-1}}}}$